Find the magnitude of the shortage distance between the lines - x - 8-3 - y - 9- - 16 - z - 10-7-x - 15-3 - y - 29-8 - z - 5- - 5

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Distance Formula

Find the magn...

Question

Find the magnitude of the shortage distance between the lines , $\frac{x - 8}{3} = \frac{y + 9}{- 16} = \frac{z - 10}{7}; \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}$

Open in App

Solution

$\begin{matrix} \frac{x - 8}{3} = \frac{y + 9}{- 16} = \frac{z - 10}{7} \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5} d = ∣ ∣ ∣ ∣ \frac{{(3 ˆ i - 16 ˆ j + 7 ˆ k) \times (3 ˆ i + 8 ˆ j - 5 ˆ k)} . (7 ˆ i + 38 ˆ j - 5 ˆ k)}{∣ ∣ (3 ˆ i - 16 ˆ j + 7 ˆ k) \times (3 ˆ i + 8 ˆ j - 5 ˆ k) ∣ ∣} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \frac{(24 ˆ i + 36 ˆ j + 72 ˆ k) \times (7 ˆ i + 38 ˆ j - 5 ˆ k)}{∣ ∣ (24 ˆ i + 36 ˆ j + 72 ˆ k) ∣ ∣} ∣ ∣ ∣ ∣ = ∣ ∣ ∣ \frac{168 + 1368 - 360}{\sqrt{576 + 1296 + 5184}} ∣ ∣ ∣ = \frac{1176}{\sqrt{7056}} = \frac{1176}{84} = 14 u n i t . \end{matrix}$

Suggest Corrections

Similar questions

Q. Find the shortest distance (in cm) between the lines

\frac{x - 8}{3} = \frac{y + 9}{- 16} = \frac{z - 10}{7}

and

\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{5 - z}{5}

Q. Find the vector equation of the line passing through point

(1, 2, - 4)

and perpendicular to the two lines

\frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7}

and

\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}

Q. The shortest distance between the lines

\frac{x - 7}{3} = \frac{y + 4}{- 16} = \frac{z - 6}{7}

and

\frac{x - 10}{3} = \frac{y - 30}{8} = \frac{4 - z}{- 5}

The equations of the line passing through the point(1,2,-4) and perpendicularto the two lines $\frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}$ , will be

Q. Find the vector and cartesian equations of the line passing through the point (1,2,-4) and perpendicular to the two lines

\frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7}

and

\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}

Join BYJU'S Learning Program

Find the magnitude of the shortage distance between the lines ,x−83=y+9−16=z−107;x−153=y−298=z−5−5

Find the magnitude of the shortage distance between the lines , $\frac{x - 8}{3} = \frac{y + 9}{- 16} = \frac{z - 10}{7}; \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}$